Solve z_{4}=2-4i/1+i

$$z _ { 4 } = \frac { 2 - 4 i } { 1 + i }$$

Solve for z_4

$$z_{4}=-1-3i$$

Steps for Solving Linear Equation

Multiply both numerator and denominator of \frac{2-4i}{1+i} by the complex conjugate of the denominator, 1-i.

$$z_{4}=\frac{\left(2-4i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$z_{4}=\frac{\left(2-4i\right)\left(1-i\right)}{1^{2}-i^{2}}$$

By definition, i^{2} is -1. Calculate the denominator.

$$z_{4}=\frac{\left(2-4i\right)\left(1-i\right)}{2}$$

Multiply complex numbers 2-4i and 1-i like you multiply binomials.

$$z_{4}=\frac{2\times 1+2\left(-i\right)-4i-4\left(-1\right)i^{2}}{2}$$

By definition, i^{2} is -1.

$$z_{4}=\frac{2\times 1+2\left(-i\right)-4i-4\left(-1\right)\left(-1\right)}{2}$$

Do the multiplications in 2\times 1+2\left(-i\right)-4i-4\left(-1\right)\left(-1\right).

$$z_{4}=\frac{2-2i-4i-4}{2}$$

Combine the real and imaginary parts in 2-2i-4i-4.

$$z_{4}=\frac{2-4+\left(-2-4\right)i}{2}$$

Do the additions in 2-4+\left(-2-4\right)i.

$$z_{4}=\frac{-2-6i}{2}$$

Divide -2-6i by 2 to get -1-3i.

$$z_{4}=-1-3i$$

Assign z_4

$$z_{4}≔-1-3i$$